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3 years ago

What is the value of the expression 3/8 ÷ 1 1/2

Mathematics

1 answer:

Shtirlitz [24]3 years ago

8 0

Answer:

0.25

Step-by-step explanation:

i used a caculator (sorry im terrible at spelling)

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3 years ago

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2 years ago

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Step-by-step explanation:

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3 years ago

In 1981, the Australian humpback whale population was 350 and has increased at a rate of 14% each year since then.)

dimulka [17.4K]

Answer:

In 1981, the Australian humpback whale population was 350

Po = Initial population = 350

rate of increase = 14% annually

P(t) = Po*(1.14)^t

P(t) = 350*(1.14)^t

Where

t = number of years that have passed since 1981

Year 2000

2000 - 1981 = 19 years

P(19) = 350*(1.14)^19

P(19) = 350*12.055

P(19) = 4219.49

P(19) ≈ 4219

Year 2018

2018 - 1981 = 37 years

P(37) = 350*(1.14)^37

P(37) = 350*127.4909

P(37) = 44621.84

P(37) ≈ 44622

There would be about 44622 humpback whales in the year 2018

5 0

3 years ago

Which is the value of this expression when a=5 and k=-2

Zinaida [17]

Answer:

Option C is correct.

Step-by-step explanation:

We are given the expression:

$(\frac{3^2a^{-2}}{3a^{-1}})^k$

The value of a =5 and k = -2

Putting the values and solving

$=(\frac{3^2*5^{-2}}{3*5^{-1}})^-2\\=(\frac{3^{2-1}}{5^{-1+2}})^-2\\=(\frac{3^{1}}{5^{1}})^-2\\\\=(\frac{3}{5})^-2\\if \,\,a^{-1} \,\,then\,\, 1/a\\=\frac{(3)^{-2}}{(5)^{-2}}\\ Can\,\,be\,\,written\,\,as\\\\=\frac{(5)^{2}}{(3)^{2}} \\=\frac{25}{9}$

Option C is correct.

3 0

3 years ago